Problem: Solve for $x$ : $ 4|x - 9| + 9 = 6|x - 9| + 6 $
Solution: Subtract $ {4|x - 9|} $ from both sides: $ \begin{eqnarray} 4|x - 9| + 9 &=& 6|x - 9| + 6 \\ \\ {- 4|x - 9|} && {- 4|x - 9|} \\ \\ 9 &=& 2|x - 9| + 6 \end{eqnarray} $ Subtract $6$ from both sides: $ \begin{eqnarray} 9 &=& 2|x - 9| + 6 \\ \\ {- 6} && {- 6} \\ \\ 3 &=& 2|x - 9| \end{eqnarray} $ Divide both sides by ${2}$ $ \dfrac{3} {{2}} = \dfrac{2|x - 9|} {{2}} $ Simplify: $ \dfrac{3}{2} = |x - 9| $ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ -\dfrac{3}{2} = x - 9 $ or $ \dfrac{3}{2} = x - 9 $ Solve for the solution where $x - 9$ is negative: $ - \dfrac{3}{2} = x - 9$ Add ${9}$ to both sides: $ \begin{eqnarray} - \dfrac{3}{2} &=& x - 9 \\ \\ {+ 9} && {+ 9} \\ \\ -\dfrac{3}{2} + 9 &=& x \end{eqnarray} $ Change the ${ + 9}$ to an equivalent fraction with a denominator of $2$ $ - \dfrac{3}{2} {+ \dfrac{18}{2}} = x $ $ \dfrac{15}{2} = x $ Then calculate the solution where $x - 9$ is positive: $ \dfrac{3}{2} = x - 9 $ Add ${9}$ to both sides: $ \begin{eqnarray} \dfrac{3}{2} &=& x - 9 \\ \\ {+ 9} && {+ 9} \\ \\ \dfrac{3}{2} + 9 &=& x \end{eqnarray} $ Change the ${ + 9}$ to an equivalent fraction with a denominator of $2$ $ \dfrac{3}{2} {+ \dfrac{18}{2}} = x $ $ \dfrac{21}{2} = x $ Thus, the correct answer is $x = \dfrac{15}{2} $ or $x = \dfrac{21}{2} $.